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part b: what is the margin of error for this interval? Banking fees have received much...

part b: what is the margin of error for this interval?
Banking fees have received much attention during the recent economic recession as banks look for ways to recover from the cri
Banking fees have received much attention during the recent economic recession as banks look for ways to recover from the crisis. A sample of 44 customers paid an average foe of $12.16 per month on their interest-bearing checking accounts. Assume the population standard deviation is $1.87. Complete parts a and b below. a. Construct a 99% confidence interval to estimate the average foo for the population The 90% confidence interval has a lower limit of s (Round to the nearest cont as needed.) and an upper limit of $
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Answer #1

Solution :

Given that,

Point estimate = sample mean = \bar x = $12.16

Population standard deviation =   \sigma = $1.87

Sample size = n =44

At 99% confidence level the z is ,

\alpha  = 1 - 99% = 1 - 0.99 = 0.01

\alpha / 2 = 0.01 / 2 = 0.005

Z\alpha/2 = Z0.005 = 2.576

Margin of error = E = Z\alpha/2* (\sigma /\sqrtn)

= 2.576 * ( 1.87/ \sqrt 44)

= 0.7

At 99% confidence interval estimate of the population mean is,

\bar x - E < \mu < \bar x + E

12.16 - 0.7 < \mu <12.16 + 0.7

11.5< \mu < 13.2

(lower limit $11.5 , upper limit$13.2)

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